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Central Automorphism Groups Fixing the Center Element-Wise

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Central Automorphism Groups Fixing the Center Element-Wise International Electronic Journal of Algebra Volume 9 (2011) 167-170 CENTRAL AUTOMORPHISM GROUPS FIXING THE CENTER ELEMENT-WISE S. Heydar Jafari Received: 27 May 2010; Revised: 23 November 2010 Communicated by Sait Hal³c³o¸glu Abstract. Let G be a ¯nite p-group. We ¯nd a necessary and su±cient condition on G such that each central automorphism of G ¯xes the center element-wise. Mathematics Subject Classi¯cation (2000): 20D45 , 20D15 Keywords: automorphisms, p-groups, nilpotent groups 1. Introduction Let G be a ¯nite group. An automorphism σ of G is said to be central if and only if it induces the identity on G=Z(G), or equivalently, g¡1σ(g) 2 Z(G) for all g 2 G, where Z(G) is the center of G. The central automorphisms of G, denoted by Autc(G), forms a normal subgroup of Aut(G), the group of automorphisms of G. Notice that Autc(G) = CAut(G)(Inn(G)). For a group H and an abelian group K, Hom(H; K) denotes the group of all homomorphisms from H to K. Let M and N be two normal subgroups of G. By AutN (G) we mean the subgroup of Aut(G) consisting of all automorphisms which induces the identity on G=N. Also by AutM (G) we mean the subgroup of Aut(G) consisting all automorphisms which N N induces identity on M. Let AutM (G) = Aut (G) \ AutM (G). we use the notation pn pn G = hg j g 2 Gi and Ct for cyclic group of order t. The group of central automorphisms of a ¯nite groups is of great importance in the investigation of Aut(G), and has been studied by several authors (see e.g., [1-7]). Z(G) In the article of Attar [2], it is proved for a ¯nite p-group G, AutZ(G)(G) = Inn(G) if and only if G is nilpotent of class 2 and Z(G) is cyclic. M.K.Yadav in [7] obtained Z(G) some necessary and su±cient conditions for the equality Autc(G) = AutZ(G)(G) on p-groups of class 2. We ¯nd a necessary and su±cient condition on G in the general case in order Z(G) for Autc(G) = AutZ(G)(G). Z(G) Theorem: Let G be a ¯nite p-group. Then Autc(G) = AutZ(G)(G) if and only 0 n 0 if Z(G)G ⊆ Gp G where exp(Z(G)) = pn . 168 S. HEYDAR JAFARI 2. Proofs Let G be a ¯nite group and M be a central subgroup of G and σ 2 AutM (G). ¡1 Then we can de¯ne a homomorphism fσ from G into M such that fσ(x) = x σ(x). On the other hand, given a homomorphism f from G to M, we can always de¯ne an endomorphism σf of G such that σf (x) = xf(x). But σf is an automorphism of G if and only if for every non-trivial element x 2 M, f(x) 6= x¡1. A ¯nite group G is said to be purely non-abelian if it has no nontrivial abelian direct factor. In [1], Adney and Yen proved that the correspondence σ 7! fσ 0 de¯ned above is a one-to-one map of Autc(G) onto Hom(G=G ;Z(G)) for purely non-abelian ¯nite groups. 0 0 We recall that if H · Z(G) and K=G is a direct factor of G=G , then any element 0 0 f of Hom(K=G ;H) induces an element f¹ of Hom(G=G ;H) which is trivial on the 0 0 complement of K=G in G=G . To simplify the notation in the proof of the main theorem, we will identify f with the corresponding homomorphism from G into H. The following Lemma can be obtained from Proposition 2.3 of [6] immediately. Lemma 2.1. [7, Lemma 2.4] Let G be a ¯nite non-abelian p-group such that Z(G) Autc(G) = AutZ(G)(G): Then G is purely non-abelian. Proof. Suppose G has a nontrivial abelian direct factor. Let G = A£H, where A is abelian and H is a non-abelian p-groups. Then Z(H) 6= 1 and so Hom(A; Z(H)) is nontrivial and by [6, Proposition 2.3] it can be assumed to be a subgroup of Autc(G). Z(G) It is clear that Hom(A; Z(H)) 6⊆ AutZ(G)(G), which is a contradiction. ¤ Lemma 2.2. Let G be a ¯nite abelian p-group and x an element of G. Then, there ps1 pst exist nontrivial elements x1; :::; xt such that G = hx1i£¢ ¢ ¢£hxti and x = x1 :::xt , where si 2 N [ f0g. r1 rt Proof. Let G = hy1i £ ¢ ¢ ¢ £ hyti and x = y1 :::yt , where ri ¸ 0. There exist si li li and si such that ri = p li and (p; li) = 1. We have hyii = hyi i and hence l1 lt G = hy1 i £ ¢ ¢ ¢ £ hyt i. This completes the proof. ¤ Z(G) Lemma 2.3. Let f 2 Hom(G; Z(G)) and σf 2 Autc(G). Then σf 2 AutZ(G)(G) if and only if Z(G) ⊆ Ker(f). Proof. This follows directly from the de¯nition of σf . ¤ 0 n 0 Lemma 2.4. Let G be a ¯nite p-group and Z(G)G ⊆ Gp G where exp(Z(G)) = pn. Then G is purely non-abelian. CENTRAL AUTOMORPHISM GROUPS FIXING THE CENTER ELEMENT-WISE 169 Proof. Let G = A £ H where A 6= 1 is abelian. Then A ⊆ Z(G) and A is not a n 0 n 0 subset of Gp G = Hp H , which is a contradiction. ¤ Proof of main theorem: By Lemmas 2.1 and 2.4 we can suppose that G is a 0 n 0 purely non-abelian p-group. Let Z(G)G ⊆ Gp G and x 2 Z(G). Then there exist 0 n elements a in G and b in G such that x = ap b. For any f 2 Hom(G; Z(G)), since 0 n exp(Z(G)) = pn and G ⊆ Ker(f), we have f(x) = f(a)p f(b) = 1:1 = 1. Hence Z(G) Z(G) ⊆ Ker(f) and σf 2 AutZ(G)(G). Z(G) 0 pn 0 Conversely, let Autc(G) = AutZ(G)(G) and suppose Z(G)G * G G . So there pn 0 exists x in Z(G) which is not in G G . By Lemma 2.2, there exist x1; :::; xt of G 0 0 0 0 ps1 0 pst 0 such that G=G = hx1G i£¢ ¢ ¢£hxtG i and xG = x1 G :::xt G . Hence for some sj n 0 p p sj n n j, xj is not in G , so p < p . Select z 2 Z(G), where O(z) = min(O(xjG ); p ), 0 0 and de¯ne f by xjG 7¡! z. Then f can be considered as a homomorphism of G=G 0 ps1 0 pst 0 into Z(G) and so σf 2 Autc(G). We obtain f(x) = f(xG ) = f(x1 G :::xt G ) = s p j psj psj f(xj ) = z : On the other hand z is a nontrivial element of Z(G). Therefore Z(G) σf is not in AutZ(G)(G), which is a contradiction. In particular it is of interest for groups of class 2 with elementary abelian centers, and generalises a result of M. J. Curran [3, Proposition 2.7]. Corollary 2.5. Let G be a ¯nite non-abelian p-group and exp(Z(G)) = p. Then Z(G) Autc(G) = AutZ(G)(G) if and only if Z(G) ⊆ Á(G), where Á(G) is frattini subgroup of G. Lemma 2.6. Let G = Cpa1 £ Cpa2 £ ¢ ¢ ¢ £ Cpat where a1 > a2 > ¢ ¢ ¢ > at and pn H · G for some integer n. If n > ak for some k 2 f1; :::; tg then, G=H and G » have equal rank, and G=H = (Cpa1 £ Cpa2 £ ¢ ¢ ¢ £ Cpak¡1 )=H £ (Cpak £ ¢ ¢ ¢ £ Cpat ). » Moreover (Cpa1 £ Cpa2 £ ¢ ¢ ¢ £ Cpak¡1 )=H = Cpn £ Cpn £ ¢ ¢ ¢ £ Cpn if and only if n H = Gp . The proof immediately follows from the fact that H \ (Cpak £ ¢ ¢ ¢ £ Cpat ) = 1. 0 Let G be a ¯nite p-group of class 2. Then G=Z(G) and G have equal exponent c p (say). Let G=Z(G) = Cpa1 £ Cpa2 £ ¢ ¢ ¢ £ Cpar ; where a1 > a2 > ¢ ¢ ¢ > ar > 0. Let k be the largest integer between 1 and r such that a1 = a2 = ¢ ¢ ¢ = ak = c. Let 0 a M = M=Z(G) = Cpa1 £ Cpa2 £ ¢ ¢ ¢ £ Cp k . Let G=G = Cpb1 £ Cpb2 £ ¢ ¢ ¢ £ Cpbs ; 0 where b1 > b2 > ¢ ¢ ¢ > bs > 0, be a cyclic decomposition of G=G such that M is 0 isomorphic to a subgroup of N = N=G = Cpb1 £Cpb2 £¢ ¢ ¢£Cpbk . Using the above terminology, we in particular obtain the main theorem of Yadav [7]. Corollary 2.7. Let G be a ¯nite non-abelian p-group of class 2. Then the following are equivalent. 170 S. HEYDAR JAFARI Z(G) (a) Autc(G) = AutZ(G)(G). n 0 (b) Z(G) = Gp G . 0 0 (c) r = s, (G=Z(G))=M =» (G=G )=N and the exponent of Z(G) and G are equal. 0 n Proof. Since exp(G=Z(G)) = exp(G ) · exp(Z(G)), we have Gp ⊆ Z(G). There- n 0 fore Gp G ⊆ Z(G), and by main theorem of article (a) and (b) are equivalent. 0 0 Let (b) hold. We ¯rst show exp(G ) = exp(Z(G)).
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